119. Prove or disprove that $A\cap B$ and $A-B$ are disjoint:
Consider that $A\cap B=\{x|x\cap A\text{ and }x\cap B\}$ and that $A-B=\{x\in A|x\notin B\} $
So, to translate this question:
“Prove or disprove that the set of elements that exist in both $A$ and $B$ shares no elements with the set of elements that exist only in the complement of $B$ relative to $A$.”
Assume: $∃ \:\times \in A \cap B$. Then, $x\in B$ and $x \notin A-B$. We conclude that $(A\cap B)\cup (A-B)=\{\}$
∴ $A\cap B$ and $A-B$ are disjoint.
This is for an introductory proofs course. I just wanted to make sure I wrote this correctly and that I understand what this is asking. Thanks for all feedback!
$\endgroup$3 Answers
$\begingroup$Your proof is fine in spirit, though it's written as if it assumes that $A\cap B$ is non-empty, which is not necessary. To be really clear, notice that two sets, $S$ and $T$ are disjoint if the following statement holds: $$x\in S\rightarrow x\not\in T.$$ So, it would be clearer to write your proof as
For any $x\in A\cap B$, it must be that $x\in B$, which implies that $x\not\in A-B$, and thus may be no element in $A\cap B$ which is also in $A-B$, so they are disjoint.
Or something to that effect. The difference being that instead of using the existential quantifier $\exists x\in A\cap B$, we basically use the universal quantifier $\forall x\in A\cap B$, which is what we want here. (When you use the existential quantifier, you're really just showing that $A\cap B$ is not a subset of $A-B$, which is not the desired property)
$\endgroup$ $\begingroup$Notice $A - B = A \cap B^c $. Hence
$$ (A \cap B) \cap (A \cap B^c) = A \cap A \cap B \cap B^c = A \cap \varnothing = \varnothing$$
$\endgroup$ $\begingroup$You can also proceed by contradiction:
Suppose there is $x \in (A \cap B) \cap (A-B)$, then
$x \in (A \cap B) \implies x \in B$
$x \in A-B \implies x \notin B$
a contradiction. It follows that $(A \cap B) \cap (A-B) = \emptyset$.
$\endgroup$
